Question

How to find the integral using limit of sum 2 maths?

Answer 1

Imagine a curve above the x-axis. The function of this graph is a continuous function defined on a closed interval [a, b], where all the values of the function are non-negative. The area bound between the curve, the points ‘x = a’ and ‘x = b’ and the x-axis is the definite integral ∫ab f(x) dx of any such continuous function ‘f’.

Definite Integral as a Limit of a Sum

Look at the following graph:

To understand this, let’s evaluate the area PRSQP between the curve y = f(x), x-axis and the coordinates ‘x = a’ and ‘x = b’. Now, divide the interval [a, b] into ‘n’ equal sub-intervals denoted as:

[x0, x1], [x1, x2], [x2, x3] …. [xn – 1, xn], where,
x0 = a, x1 = a + h, x2 = a + 2h, x3 = a + 3h ….. xr = a + rh and xn = b = a + nh
Or, n = (b – a)/h. Note that as n → ∞, h → 0.

Now, the region PRSQP under consideration is the sum of all the ‘n’ sub-regions, where each sub-region is defined on subintervals [xr – 1, xr], r = 1, 2, 3 … n. Now, look at the region ABDM in the figure above. We can make the following observation:

Area of the rectangle (ABLC) < Area of the region (ABDCA) < Area of the rectangle (ABDM)        (1)

Also, note that as, h → 0 or xr – xr – 1 → 0, all these three areas become nearly equal to each other. Hence, we have

sn = h [f(x0) + f(x1) + f(x2) + …. f(xn – 1)] = h r=0∑n–1 f(xr) … (2)
and, Sn = h [f(x1) + f(x2) + f(x3) + …. f(xn)] = h r=1∑n f(xr) … (3)

Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively. To bring it into perspective, equation (1) can be re-written as:

sn < area of the region (PRSQP) < Sn … (4)

As n → ∞, these strips become narrower

Further, it is assumed that the limiting values of (2) and (3) are the same in both cases and the common limiting value is the required area under the curve. Symbolically, we have

limn → ∞ Sn = limn → ∞ sn = area of the region (PRSQP) = ∫ab f(x) dx … (5)

This area is also the limiting value of any area which is between that of the rectangles below the curve and that of the rectangles above the curve. For convenience, we shall take the rectangles having height equal to that of the curve at the left-hand-edge of each sub-interval. Hence, equation (%) is re-written as:

∫ab f(x) dx = limn → ∞ h [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Or, ∫ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)] … (6)

Where, h = (b – a)/n → 0 as n → ∞. This equation is the definition of Definite Integral as the limit of a sum.

Note: The value of the definite integral of a function over any particular interval depends on the function and the interval, but not on the variable of integration that we choose to represent the independent variable. Hence, the variable of integration is called a dummy variable.

Example 1

Find ∫02 (x2 + 1) dx as the limit of a sum.

Solution: From equation (6) above, we know that
∫ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Where, h = (b – a)/n

In this example, we have a = 0, b = 2, f(x) = (x2 + 1) and h = (2 – 0)/n = 2/n. Therefore,

∫02 (x2 + 1) dx = 2 limn → ∞ (1/n) [f(0) + f(2/n) + f(4/n) + …. + f(2{n – 1}/n)]
= 2 limn → ∞ (1/n) [1 + {(22/n2) + 1} + {(42/n2) + 1} + …. + {(2n – 2)2/n2 + 1}]
= 2 limn → ∞ (1/n) [1 + 1 + 1 + … + 1(n-times)] + 1/n2) [22 + 42 + … (2n – 2)2]
= 2 limn → ∞ (1/n) [n + 22/n2 (12 + 22 + … (n – 1)2]
= 2 limn → ∞ (1/n) [n + 4/n2 {(n – 1) n (2n – 1) / 6}]
= 2 limn → ∞ (1/n) [n + 2/3 {(n – 1) (2n – 1) / n}]
= 2 limn → ∞ (1/n) [n + 2/3 (1 – 1/n) (2 – 1/n)]
As n → ∞, 1/n → 0. Therefore, we have
∫02 (x2 + 1) dx = 2 [1 + 4/3] = 14/3.

it's not copied... it's technically written by me...

Mark it as brainliest... plz...

I hope it will help you dear...

Answer 2

Imagine a curve above the x-axis. The function of this graph is a continuous function defined on a closed interval [a, b], where all the values of the function are non-negative. The area bound between the curve, the points ‘x = a’ and ‘x = b’ and the x-axis is the definite integral ∫ab f(x) dx of any such continuous function ‘f’.

Definite Integral as a Limit of a Sum

Look at the following graph:

To understand this, let’s evaluate the area PRSQP between the curve y = f(x), x-axis and the coordinates ‘x = a’ and ‘x = b’. Now, divide the interval [a, b] into ‘n’ equal sub-intervals denoted as:

[x0, x1], [x1, x2], [x2, x3] …. [xn – 1, xn], where,
x0 = a, x1 = a + h, x2 = a + 2h, x3 = a + 3h ….. xr = a + rh and xn = b = a + nh
Or, n = (b – a)/h. Note that as n → ∞, h → 0.

Now, the region PRSQP under consideration is the sum of all the ‘n’ sub-regions, where each sub-region is defined on subintervals [xr – 1, xr], r = 1, 2, 3 … n. Now, look at the region ABDM in the figure above. We can make the following observation:

Area of the rectangle (ABLC) < Area of the region (ABDCA) < Area of the rectangle (ABDM)        (1)

Also, note that as, h → 0 or xr – xr – 1 → 0, all these three areas become nearly equal to each other. Hence, we have

sn = h [f(x0) + f(x1) + f(x2) + …. f(xn – 1)] = h r=0∑n–1 f(xr) … (2)
and, Sn = h [f(x1) + f(x2) + f(x3) + …. f(xn)] = h r=1∑n f(xr) … (3)

Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively. To bring it into perspective, equation (1) can be re-written as:

sn < area of the region (PRSQP) < Sn … (4)

As n → ∞, these strips become narrower

Further, it is assumed that the limiting values of (2) and (3) are the same in both cases and the common limiting value is the required area under the curve. Symbolically, we have

limn → ∞ Sn = limn → ∞ sn = area of the region (PRSQP) = ∫ab f(x) dx … (5)

This area is also the limiting value of any area which is between that of the rectangles below the curve and that of the rectangles above the curve. For convenience, we shall take the rectangles having height equal to that of the curve at the left-hand-edge of each sub-interval. Hence, equation (%) is re-written as:

∫ab f(x) dx = limn → ∞ h [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Or, ∫ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)] … (6)

Where, h = (b – a)/n → 0 as n → ∞. This equation is the definition of Definite Integral as the limit of a sum.

Note: The value of the definite integral of a function over any particular interval depends on the function and the interval, but not on the variable of integration that we choose to represent the independent variable. Hence, the variable of integration is called a dummy variable.

Example 1

Find ∫02 (x2 + 1) dx as the limit of a sum.

Solution: From equation (6) above, we know that
∫ab f(x) dx = (b – a) limn → ∞ (1/n) [f(a) + f(a + h) + …. + f(a + {n – 1}h)]
Where, h = (b – a)/n

In this example, we have a = 0, b = 2, f(x) = (x2 + 1) and h = (2 – 0)/n = 2/n. Therefore,

∫02 (x2 + 1) dx = 2 limn → ∞ (1/n) [f(0) + f(2/n) + f(4/n) + …. + f(2{n – 1}/n)]